24 November 2025 · Shift 3

Discount and selling price Discount is always reckoned on the marked price, so selling price = marked price $\times \frac{100-\mathrm{discount}\%}{100}$. $SP=800\times \frac{100-37.5}{100}$ $=800\times \frac{62.5}{100}=800\times \frac{5}{8}$ =500 So the selling price is Rs. 500. Hence the correct option is (d).

Discount amount The discount amount is the given percentage of the marked price. Discount $=1500\times \frac{20}{100}$ $=15\times 20=300$ So the buyer saves Rs. 300 and pays Rs. 1,200. Hence the correct option is (a).

Perimeter of a parallelogram Opposite sides of a parallelogram are equal, so the perimeter is twice the sum of two adjacent sides. 2[(2x-4)+(3x+2)]=38 2(5x-2)=38 5x-2=19 $5x=21\Rightarrow x=\frac{21}{5}=4.2$ Hence the correct option is (a).

Cone - volume and slant height Volume of a cone $=\frac{1}{3}\pi r^{2}h$; use it first to get the vertical height. $1232=\frac{1}{3}\times \frac{22}{7}\times 7^2\times h$ $1232=\frac{154}{3}h\Rightarrow h=\frac{1232\times 3}{154}=24$ So the height is 24 cm. Slant height $l=\sqrt{r^2+h^2}=\sqrt{7^2+{24}^2}=\sqrt{49+576}=\sqrt{625}=25$ Hence the correct option is (b).

Rational numbers Between any two different rational numbers there always lies one more rational number, for instance their average. $\frac{1+2}{2}=\frac{3}{2}$ lies between 1 and 2. $\frac{1+\frac{3}{2}}{2}=\frac{5}{4}$ lies between 1 and $\frac{3}{2}$, and the same step can be repeated for ever. So the list never comes to an end and the count cannot be finite. Hence the correct option is (b).

Angles of a parallelogram Two adjacent angles of a parallelogram are co-interior angles between parallel sides, so they are supplementary. Let the two angles be 10x and 8x. $10x+8x={180}^{\circ }\Rightarrow 18x={180}^{\circ }\Rightarrow x={10}^{\circ }$ Smaller angle $=8x=8\times {10}^{\circ }={80}^{\circ }$ Hence the correct option is (a).

Inradius of a triangle For every triangle, area $=r\times s$, where r is the inradius and s the semi-perimeter. $110=r\times 22$ $r=\frac{110}{22}=5$ So the circle that fits exactly inside the triangle has radius 5 cm. Hence the correct option is (c).

Volume of a pyramid The base is an equilateral triangle of side 8 cm, so its area $=\frac{\sqrt{3}}{4}{a}^2$. $=\frac{\sqrt{3}}{4}\times 8^2=\frac{\sqrt{3}}{4}\times 64=16\sqrt{3}$ sq cm Volume of a pyramid $=\frac{1}{3}\times$ base area $\times$ height. $=\frac{1}{3}\times 16\sqrt{3}\times 15=16\sqrt{3}\times 5$ $=80\sqrt{3}$ cubic cm Hence the correct option is (b).

Profit percentage Let the selling price of one article be s and its cost price be c. Gain on 14 articles =14s-14c, and the question says this gain equals the selling price of 4 articles, i.e. 4s. 14s-14c=4s $10s=14c\Rightarrow \frac{s}{c}=\frac{7}{5}$ So taking c=5 gives s=7, and the profit on one article is 7-5=2. Profit $\%=\frac{2}{5}\times 100=40\%$ Hence the correct option is (a).

Compound interest - broken period When interest is compounded yearly and the time has an odd half year, that half year earns simple interest at half the rate, i.e. $20\%$. Amount $=P{\left(1+\frac{40}{100}\right)}^2\left(1+\frac{20}{100}\right)=P\times \frac{7}{5}\times \frac{7}{5}\times \frac{6}{5}$ $=\frac{294}{125}P$ Compound interest = amount - principal $=\frac{294}{125}P-P=\frac{169}{125}P$ $\frac{169}{125}P=6760$ $P=\frac{6760\times 125}{169}=40\times 125=5000$ So the principal is ₹5,000. Hence the correct option is (c).

LCM of numbers The least number that is divisible by each of several numbers is their LCM. 7, 11 and 13 are all prime numbers, so no two of them share a common factor. For such numbers the LCM is simply their product. $\mathrm{LCM}\left(7,11,13\right)=7\times 11\times 13=77\times 13=1001$ Hence the correct option is (c).

Weighted average Number of females $=60\%$ of $800=\frac{60}{100}\times 800=480$ Number of males =800-480=320 Total height of all members $=320\times 150+480\times 170=48000+81600=129600$ cm Average height $=\frac{\mathrm{total}\mathrm{height}}{\mathrm{total}\mathrm{members}}=\frac{129600}{800}=162$ cm Hence the correct option is (d).

Relative speed The two buses move towards each other, so the gap between them closes at the sum of their speeds. Relative speed =40+80=120 km/h Time to meet $=\frac{\mathrm{distance}\mathrm{between}\mathrm{them}}{\mathrm{relative}\mathrm{speed}}=\frac{600}{120}=5$ hours Hence the correct option is (b).

Partnership profit share All three invest for the same period of one year, so the profit is divided in the ratio of the investments alone. Ratio =150000:80000:120000=15:8:12 Sum of the ratio terms =15+8+12=35 One part $=\frac{105000}{35}=3000$ Beena's share $=8\times 3000=24000$ Hence the correct option is (c).

Square root 8464 ends in the digit 4, so its square root must end in 2 or 8 (since $2^2=4$ and $8^2=64$). ${90}^2=8100$ and ${95}^2=9025$, so the root lies between 90 and 95; only 92 from that range fits the last-digit test. Check: ${92}^2={\left(90+2\right)}^2=8100+2\times 90\times 2+4=8100+360+4=8464$ So $\sqrt{8464}=92$ Hence the correct option is (d).

Mixture and replacement Milk $=\frac{7}{10}\times 80=56$ litres and water $=\frac{3}{10}\times 80=24$ litres. The 16 litres taken out is $\frac{16}{80}=\frac{1}{5}$ of the mixture, so $\frac{4}{5}$ of each liquid is left. Milk left $=\frac{4}{5}\times 56=44.8$ litres; water left $=\frac{4}{5}\times 24=19.2$ litres. Pure milk is added, so only the milk increases: milk =44.8+16=60.8 litres, water stays 19.2 litres. New ratio =60.8:19.2=608:192=19:6 Hence the correct option is (a).

Incentre angle First find the third angle: $\angle C={180}^{\circ }-\angle A-\angle B={180}^{\circ }-{50}^{\circ }-{80}^{\circ }={50}^{\circ }$ Since I is the incentre, AI bisects $\angle A$ and BI bisects $\angle B$. In $△AIB$: $\angle IAB=\frac{{50}^{\circ }}{2}={25}^{\circ }$ and $\angle IBA=\frac{{80}^{\circ }}{2}={40}^{\circ }$ $\angle AIB={180}^{\circ }-{25}^{\circ }-{40}^{\circ }={115}^{\circ }$ The same result comes from the standard formula $\angle AIB={90}^{\circ }+\frac{\angle C}{2}={90}^{\circ }+{25}^{\circ }={115}^{\circ }$. Hence the correct option is (c).

Perpendicular lines slope Write the given line in the form y=mx+c: $3x+4y=16\Rightarrow 4y=-3x+16\Rightarrow y=-\frac{3}{4}x+4$ So its slope is $m_1=-\frac{3}{4}$. Two lines are perpendicular when $m_1\times m_2=-1$, so the required slope is $m_2=-\frac{1}{m_1}=\frac{4}{3}$. Option (c) is $y=\frac{4x}{3}-3$, whose slope is $\frac{4}{3}$; check: $-\frac{3}{4}\times \frac{4}{3}=-1$. Options (a) and (d) both have slope $\frac{3}{4}$ and option (b) has slope $-\frac{3}{5}$, so none of them gives -1. Hence the correct option is (c).

Length of tangent Let O be the centre of the larger circle (radius OT=10 cm) and P the centre of the smaller circle, so OP=15 cm. The tangent drawn from P touches the larger circle at T, and a radius drawn to the point of contact is perpendicular to the tangent, so $\angle OTP={90}^{\circ }$. By Pythagoras in the right triangle OTP: $P{T}^2=O{P}^2-O{T}^2$ $P{T}^2={15}^2-{10}^2=225-100=125$ $PT=\sqrt{125}$ cm The radius 3 cm of the smaller circle is not needed here, because the tangent starts at its centre and not on its circumference. Hence the correct option is (c).

Percentage relations Q=500, and P is 20 percent more than Q: $P=500\times \frac{120}{100}=600$ So P+Q=600+500=1100 X+Y is 100 percent more than P+Q, i.e. double it: $X+Y=2\times 1100=2200$ X is 200 percent more than Y means $X=Y+\frac{200}{100}Y=3Y$ $3Y+Y=2200\Rightarrow 4Y=2200\Rightarrow Y=550$ $X=3\times 550=1650$ Hence the correct option is (b).

Compound interest sum For compound interest, Amount $=P{\left(1+\frac{r}{100}\right)}^n$. Dividing the 4-year amount by the 2-year amount cancels P: ${\left(1+\frac{r}{100}\right)}^2=\frac{12884.08}{10648}=1.21$ $1+\frac{r}{100}=\sqrt{1.21}=1.1$, so $r=10\%$. Now use the 2-year amount: $P\times {\left(1.1\right)}^2=10648\Rightarrow P\times 1.21=10648$ $P=\frac{10648}{1.21}=8800$ Hence the correct option is (c).

Simple interest sum Simple interest is $\frac{P\times R\times T}{100}$, so the interest for 12 years minus the interest for 8 years is simply the interest for 12-8=4 years. $\frac{P\times 10\times 4}{100}=80$ $\frac{40P}{100}=80\Rightarrow \frac{2P}{5}=80$ $P=\frac{80\times 5}{2}=200$ Hence the correct option is (a).

Discount and marked price Let the discount be 100. The selling price is 150 percent more than the discount, so $SP=100+\frac{150}{100}\times 100=100+150=250$ Marked price = selling price + discount =250+100=350 Required percentage $=\frac{SP}{MP}\times 100=\frac{250}{350}\times 100=\frac{500}{7}=71.42\%$ Hence the correct option is (a).

Time and work Take the whole work as 24 units, so A and B together do $\frac{24}{24}=1$ unit per day. In the first 8 days they finish $8\times 1=8$ units. Remaining work =24-8=16 units, and B alone does this in 20 days. B's one day's work $=\frac{16}{20}=0.8$ unit. A's one day's work =1-0.8=0.2 unit. Days needed by A alone $=\frac{24}{0.2}=120$ Hence the correct option is (b).

Work on alternate days Take the whole work as the LCM of 24 and 48, i.e. 48 units. P does $\frac{48}{24}=2$ units a day and Q does $\frac{48}{48}=1$ unit a day. Working alternately with P starting, one pair of days finishes 2+1=3 units. Number of such pairs needed $=\frac{48}{3}=16$ Total days $=16\times 2=32$ The work ends exactly at the close of a Q-day, so no part day has to be adjusted. Hence the correct option is (b).